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If Tan − 1 ( X 2 − Y 2 X 2 + Y 2 ) = a Prove that D Y D X = X Y ( 1 − Tan a ) ( 1 + Tan a ) ? - Mathematics

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प्रश्न

If \[\tan^{- 1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = a\] Prove that  \[\frac{dy}{dx} = \frac{x}{y}\frac{\left( 1 - \tan a \right)}{\left( 1 + \tan a \right)}\] ?

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उत्तर

\[\text{ We have }, \tan^{- 1} \left( \frac{x^2 - y^2}{x^2 + y^2} \right) = a\]

\[ \Rightarrow \frac{x^2 - y^2}{x^2 + y^2} = \tan a\]

\[ \Rightarrow x^2 - y^2 = \tan a\left( x^2 + y^2 \right)\] 

Differentiating with respect to x,

\[\Rightarrow \frac{d}{dx}\left( x^2 - y^2 \right) = \tan a\frac{d}{dx}\left( x^2 + y^2 \right)\]

\[ \Rightarrow \left( 2x - 2y\frac{d y}{d x} \right) = \tan a\left( 2x + 2y\frac{d y}{d x} \right)\]

\[ \Rightarrow 2x - 2y\frac{d y}{d x} = 2x\tan a + 2y\tan a\frac{d y}{d x}\]

\[ \Rightarrow 2y\tan a\frac{d y}{d x} + 2y\frac{d y}{d x} = 2x - 2x\tan a\]

\[ \Rightarrow 2y\left( 1 + \tan a \right)\frac{d y}{d x} = 2x\left( 1 - \tan a \right)\]

\[ \Rightarrow \frac{d y}{d x} = \frac{x}{y}\left( \frac{1 - \tan a}{1 + \tan a} \right)\]

Hence proved

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पाठ 11: Differentiation - Exercise 11.04 [पृष्ठ ७५]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 11 Differentiation
Exercise 11.04 | Q 19 | पृष्ठ ७५

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